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15 Chemical Kinetics

15 Chemical Kinetics. Chemical kinetics : the study of reaction rate ,  a quantity conditions affecting it, the molecular events during a chemical reaction (mechanism), and presence of other components (catalysis).

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15 Chemical Kinetics

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  1. 15 Chemical Kinetics Chemical kinetics: the study of reaction rate,  a quantity conditions affecting it,the molecular events during a chemical reaction (mechanism), and presence of other components (catalysis). Factors affecting reaction rate:Concentrations of reactants Catalyst Temperature Surface area of solid reactants or catalyst What quantities do we study regarding chemical reactions? 15 Chemical Kinetics

  2. Reaction Rate Defined Reaction rate: changes in a concentration of a product or a reactant per unit time. [ ] concentrationReaction rate = —— t [ ] change [ ] t Define reaction rate and explainAverage reaction rateInstantaneous reaction rate(2 tangents shown)Initial reaction rate t 15 Chemical Kinetics

  3. Expressing reaction rates For a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation. Note the expression and reasons for their relations for the reaction 2 NO + O2 (g) = 2 NO2 (g) [O2] 1 [NO] 1 [NO2] Reaction rate = –——— = –————— = ————  t 2  t 2  t Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for 2 N2O5 = 4 NO2 + O2 How can the rate expression be unique and universal? 15 Chemical Kinetics

  4. Calculating reaction rate The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for 2 N2O5 = 4 NO2 + O2 Solution: (0.93 – 1.24)e-2 – 0.31e-2 MDecomposition rate of N2O5 = – ———————— = – —————— 1200 – 600 600 s = 5.2e-6 M s-1. Note however, rate of formation of NO2 = 1.02e-5 M s-1. rate of formation of O2 = 2.6e-6 M s-1. The reaction rates are expressed in 3 forms Be able to do this type problems 15 Chemical Kinetics

  5. Determine Reaction Rates To measure reaction rate, we measure the concentration of either a reactant or product at several time intervals. The concentrations are measured using spectroscopic method or pressure (for a gas). For example, the total pressure increases for the reaction: 2 N2O5 (g)  4 NO2 (g) + O2(g) Because 5 moles of gas products are produced from 2 moles of gas reactants. For the reaction CaCO3 (s) CaO(s) + CO2 (g) The increase in gas pressure is entirely due to CO2 formed. barometer 15 Chemical Kinetics

  6. Differential Rate Laws Dependence of reaction rate on the concentrations of reactants is called the rate law, which is unique for each reaction. For a general reaction, a A + b B + c C  products the rate law has the general form order wrt A, B, and C, determined experimentally reaction rate = k [A]X [B]Y [C]Zthe rate constant For example, the rate law is rate = k [Br-] [BrO3-] [H+] for 5 Br- + BrO3- + 6 H+ 3Br2 + 3 H2O The reaction is 1storder wrt all three reactants, total order 3. Use differentials to express rates 15 Chemical Kinetics

  7. Variation of Reaction rates and Order 2nd order, rate = k [A]2 rate First order, rate = k [A] k = rate, 0th order [A] [A] = ___? The variation of reaction rates as functions of concentration for various order is interesting. Mathematical analysis is an important scientific tool, worth noticing. 15 Chemical Kinetics

  8. Differential Rate Law determination Estimate the orders and rate constant k from the results observed for the reaction? What is the rate when [H2O2] = [I-] = [H+] = 1.0 M? H2O2 + 3 I- + 2 H+  I3- + 2 H2O Exprmt [H2O2] [I-] [H+] Initial rate M s-1 1 0.010 0.010 0.0050 1.15e-6 2 0.020 0.010 0.0050 2.30e-6 3 0.010 0.020 0.0050 2.30e-6 4 0.010 0.010 0.0100 1.15e-6 Learn the strategy to determine the rate law from this example. Figure out the answer without writing down anything. Solution next 15 Chemical Kinetics

  9. Differential Rate Law determination - continue Estimate the orders from the results observed for the reaction H2O2 + 3 I- + 2 H+  I3- + 2 H2O Exprmt [H2O2] [I-] [H+] Initial rate M s-1 1 0.010 0.010 0.0050 1.15e-6 2 0.020 0.010 0.0050 2.30e-6 1 for H2O2 3 0.010 0.020 0.0050 2.30e-6 1 for I- 4 0.010 0.010 0.0100 1.15e-6 0 for H+ 1.15e-6 = k[H2O2]x [I-]y [H+]z 1.15e-6 k (0.010)x(0.010)y(0.0050)z  exprmt 1 1 1 ----------- = ------------------------------------- ---- = --- 2.30e-6 k (0.020)x(0.010)y(0.0050)z exprmt 2 2 2 x = 1 ( )x 15 Chemical Kinetics

  10. Differential Rate Law determination - continue Other orders are determined in a similar way as shown before.Now, lets find k and the rate Thurs, rate = 1.15e-6 = k (0.010)(0.010) from exprmt 1 k = 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1 And the rate law is therefore, –d [H2O2] k rate = ————— = 0.0115[H2O2] [I-] a differential rate lawd ttotal order 2 The rate when [H2O2] = [I-] = [H+] = 1.0 M: The rate is the same as the rate constant k, when concentrations of reactants are all unity (exactly 1), doesn’t matter what the orders are. 15 Chemical Kinetics

  11. Differential Rate Law determination – continue The reaction rate –d[H2O2]/dt = 0.0115 [H2O2] [I–], for H2O2 + 3 I- + 2 H+  I3- + 2 H2OWhat is – d[I–]/dt when [H2O2] = [I–] = 0.5? Solution: Please note the stoichiometry of equation and how the rate changes. – d[I–]/dt = – 3 d[H2O2]/dt = 3* 0.0115 [H2O2] [I–] = 0.0345 * 0.5 * 0.5 = 0.0086 M s-1 In order to get a unique rate constant k, we evaluate k for the reaction a A + b B product this way rate = -1/ad[A]/dt = -1/bd[B]/dt = k [A]x [B]y Note the reaction rate expression and the stoichiometry of equation. 15 Chemical Kinetics

  12. Differential Rate Law determination - continue From the following reaction rates observed in 4 experiments, derive the rate law for the reaction A + B + C productswhere reaction rates are measured as soon as the reactants are mixed. Expt 1 2 3 4 [A]o0.100 0.200 0.200 0.100 [B]o0.100 0.100 0.300 0.100 [C]o0.100 0.100 0.100 0.400rate0.100 0.800 7.200 0.400 This example illustrates the strategy to determine, and a reliable method to solve rate-law experimentally. Solution next 15 Chemical Kinetics

  13. Differential Rate Law determination - continue From the following reaction rates, derive the rate law for the reaction A + B + C productswhere reaction rates are measured as soon as the reactants are mixed. Expt 1 2 3 4 order [A]o 0.100 0.200 0.200 0.100 3 from expt 1 & 2 [B]o 0.100 0.100 0.300 0.100 2expt 1, 2 & 3 [C]o 0.100 0.100 0.100 0.400 1expt 1 & 4rate 0.100 0.800 7.200 0.400 Assume rate = k [A]x[B]y[C]z 0.800 k 0.2x 0.1y 0.1z ----- = ----------------------0.100 k 0.1x 0.1y 0.1z Therefore 8 = 2xlog 8 = x log 2x = log 8 / log 2 = 3 15 Chemical Kinetics

  14. Integrated Rate Laws concentrations as functions of time One reactant A decomposes in 1st or 2nd order rate law. Differential rate law Integrated rate law – d[A] / dt = k [A] = [A]o–k t d[A]– —— = k [A] [A] = [A]o e – k t or ln [A] = ln [A]o–k td t d[A] 1 1 [A] conc at t– —— = k [A]2 —— – —— = k td t [A] [A]o[A]o conc at t = 0 Describe, derive and apply the integrated rate laws Learn the strategy to determine rate-law 15 Chemical Kinetics

  15. Concentration and time of 1st order reaction Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times. [A] ln[A] ln [A] = ln [A]o – k t [A] = [A]o e – k t t½ t t 15 Chemical Kinetics

  16. Half life & k of First Order Decomposition The time required for half of A to decompose is called half life t1/2. Since [A] = [A]o e – k t or ln [A] = ln [A]o – k t When t = t1/2, [A] = ½ [A]o Thus ln ½ [A]o = ln [A]o–k t1/2 – ln 2 = –k t1/2 k t1/2 = ln 2 = 0.693  relationship between k and t1/2 Radioactive decay usually follow 1st order kinetics, and half life of an isotope is used to indicate its stability. Evaluate t½ from k or k from t½ 15 Chemical Kinetics

  17. 1st order reaction calculation N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½and percent decomposed in 500 s. If the rate-law is known, what are the key parameters? Solution next 15 Chemical Kinetics

  18. 1st order reaction calculation N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½and percent decomposed in 500 s. Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or 0.9 = 1.0 e – k tapply [A]o = [A] e– k t ln 0.9 = ln 1.0 – k 30 s – 0.1054 = 0 – k * 30k = 0.00351 s – 1t½ = 0.693 / k = 197 s apply k t ½ = ln 2 [A] = 1.0 e – 0.00351*500 = 0.173Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 % After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed.After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed. Apply integrated rate law to solve problems 15 Chemical Kinetics

  19. Typical Problem wrt 1st Order Reaction The decomposition of A is first order, and [A] is monitored. The following data are recorded: t / min 0 2 4 8[A]/[M] 0.100 0.0905 0.0819 0.0670 Calculate k (What is the rate constant? k = 0.0499)Calculate the half life (What is the half life? Half life = 13.89)Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?)Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A to decompose.) Work out all the answers 15 Chemical Kinetics

  20. A 2nd Order Example Dimerization of butadiene is second order: 2 C4H6(g) = C8H12(g). The rate constantk at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 2.0 M. Calculate the time required for [B] = 1.0 and 0.5 M Calculate concentration of butadiene when t = 1, 5, 10, and 30. If the rate-law is known, what are the key parameters? Apply the right model and work out all the parameters. Solution next 15 Chemical Kinetics

  21. 1 1 —— – —— = k t [B] [B]o 1 1 —— – —— [B] [B]ot = ————————k [B]o [B] = —————— [B]ok t + 1 A 2nd Order Example Dimerization of butadiene is second order: 2 C4H6(g) = C8H12(g). The rate constant k at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 2.0 M. Calculate the time t required for [B] = 1.0 and 0.5 M Calculate concentration of butadiene when t = 1, 5, 10, and 30. t = 1 5 10 15 30 35 [B] = 1.67 1.0 0.67 0.50 0.29 0.25 Work out the formulas and then evaluate values 15 Chemical Kinetics

  22. Half life of 2nd Order Chemical Kinetics 1/[B] t = 1 5 10 15 30 35 [B] = 1.67 1.0 0.67 0.50 0.29 0.25 How does half life vary in 2nd order reactions? 15 Chemical Kinetics

  23. Plot of [B] vs. t & 1/[B] vs. t for 2nd Order Reactions t = 1 5 10 15 30 35 [B] = 1.67 1.0 0.67 0.50 0.29 0.25 [B] 1— [B] [B]o [B] = —————— [B]ok t + 1 1 1 —— – —— = k t [B] [B]o t t What kind of plot is linear for 1st and 2nd reactions? 15 Chemical Kinetics

  24. Chemical Reaction and Molecular Collision Molecular collisions lead to chemical reactions. Thus, the reaction constant, k is determined by several factors. k = Z f p Z: collision frequencyp, the fraction with proper orientationf, fraction of collision having sufficient energy for reactionf is related to the potential energy barrier called activation energy, Ea. f e – Ea / RTor exp (– Ea/ R T) Thus, k = Ae – Ea / RT constant Potential energy constant Ea reaction How does temperature affect reaction rates? Explain energy aspect in a chemical reaction 15 Chemical Kinetics

  25. Energy in chemical reactions Potential energy RA-PDactivated complex R + A Ea Eafor reverse reaction H exothermic  P + D Endothermic rxn Progress of reaction Explain the various terms and energy changes in a chemical reaction 15 Chemical Kinetics

  26. The Arrhenius Equation The temperature dependence of the rate constant k is best described by the Arrhenius equation: k = A e – Ea / R Tor ln k = ln A – Ea / R T If k1 and k2 are the rate constants at T1 and T2 respectively, then k1Ea1 1 ln —— = –— — – — k2 R T1 T2 1903 Nobel Prize citation” …in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation” How does temperature affect reaction rates? Derive and apply these relationship to solve problems, and recall the Clausius-Clapeyron equation. 15 Chemical Kinetics

  27. Application of Arrhenius Equation From k = A e – Ea / R T, calculate A, Ea, k at a specific temperature and T. The reaction: 2 NO2(g) -----> 2NO(g) + O2(g) The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11? Method: derive various versions of the same formula k = A e – Ea / R T A= k e Ea / R T A / k = e Ea / R T ln(A / k) = Ea / R T Make sure you know how to transform the formula into these forms. 15 Chemical Kinetics

  28. Application of Arrhenius Equation (continue) The reaction: 2 NO2(g) -----> 2NO(g) + O2(g) The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11? Use the formula derived earlier: A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1 K –1*300 K)) = 2.13e9 s-1 k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K) = 1.23e-12 s-1 T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1K-1 = 285 K 15 Chemical Kinetics

  29. The Effect of Temperature on Reaction Rates Reaction rate = k [A}x[B]y[C]z (Concentration effect at constant T) k = A exp ( – Ea / RT) (Temperature effect) Use graphic method to discuss the variation of k vs. Tvariation of k vs 1 / Tvariation of ln(k) vs Tvariation of ln(k) vs 1 / T See a potential multiple choice question in an exam? 15 Chemical Kinetics

  30. Elementary Reactions and Mechanism Elementary reactions are steps of molecular events showing how reactions proceed. This type of description is a mechanism. The mechanism for the reaction between CO and NO2 is proposed to be Step 1 NO2 + NO2 NO3 + NO (an elementary reaction) Step 2 NO3 + CO  NO2 + CO2 (an elementary reaction) Add these two equations led to the overall reaction NO2 + CO = NO + CO2 (overall reaction) A mechanism is a proposal to explain the rate law, and it has to satisfy the rate law. A satisfactory explanation is not a proof. Explain terms in red 15 Chemical Kinetics

  31. Molecularity of Elementary Reactions The total order of rate law in an elementary reaction is molecularity. The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react. A molecule decomposes by itself is a unimolecular reaction (step);two molecules collide and react is a bimolecular reaction (step); &three molecules collide and react is a termolecular reaction (step). O3 O2 + O rate = k [O3] NO2 + NO2  NO3 + NO rate = k [NO2]2 Br + Br + Ar  Br2 + Ar* rate = k [Br]2[Ar] Caution: Derive rate laws this way only for elementary reactions. 15 Chemical Kinetics

  32. Molecularity of elementary reactions - Example Some elementary reactions for the reaction between CH4 and Cl2 are Cl2 2 Cl 2 Cl  Cl2 2Cl + CH4  Cl2 + CH4* Cl + CH4  HCl + CH3 CH3 + Cl  CH3Cl CH3 + CH3  CH3-CH3 CH3Cl + Cl  HCl + CH2Cl CH2Cl + Cl  CH2Cl2 * * * (more) Write down the rate laws and describe them as uni- bi- or ter-molecular steps yourself, please. 15 Chemical Kinetics

  33. NO + O2 + NO2 NO2 + NO3 Elementary Reactions are Molecular Events N2O5 NO2 + NO3  Explain differences between elementary and over reaction equations 15 Chemical Kinetics

  34. Rate Laws and Mechanisms A mechanism is a collection of elementary steps devise to explain the the reaction in view of the observed rate law. You need the skill to derive a rate law from a mechanism, but proposing a mechanism is task after you have learned more chemistry For the reaction, 2 NO2 (g) + F2 (g)  2 NO2F (g), the rate law is, rate = k [NO2] [F2] . Can the elementary reaction be the same as the overall reaction? If they were the same the rate law would have been rate = k [NO2]2 [F2], Therefore, they the overall reaction is not an elementary reaction. Its mechanism is proposed next. 15 Chemical Kinetics

  35. Rate-determining Step in a Mechanism The rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction. The (determined) rate law is, rate = k [NO2] [F2],for the reaction, 2 NO2 (g) + F2 (g)  2 NO2F (g), and a two-step mechanism is proposed: i NO2 (g) + F2 (g)  NO2F (g) + F (g) ii NO2 (g) + F (g)  NO2F (g)Which is the rate determining step? Answer:The rate for step i is rate = k [NO2] [F2], which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step. Explain rate determining step in a mechanism and use it to derive the rate law. 15 Chemical Kinetics

  36. Deriving a Rate Law From a Mechanism - 0 The decomposition of H2O2 in the presence of I– follow this mechanism, i H2O2 + I–  k1® H2O + IO– slow ii H2O2 + IO–  k2® H2O + O2 + I– fast What is the rate law? Energy Eai Eaii reaction Solve the problem 15 Chemical Kinetics

  37. Deriving a rate law from a mechanism - 1 The decomposition of H2O2 in the presence of I– follow this mechanism, i H2O2 + I–  k1® H2O + IO– slow ii H2O2 + IO–  k2® H2O + O2 + I– fast What is the rate law? Solution The slow step determines the rate, and the rate law is: rate = k1 [H2O2] [I –] Since both [H2O2] and [I –]are measurable in the system, this is the rate law. 15 Chemical Kinetics

  38. Deriving a rate law from a mechanism - 2 Derive the rate law for the reaction, H2 + Br2 = 2 HBr, from the proposed mechanism: i Br2 2 Br fast equilibrium (k1, k-1)ii H2 + Br  k2® HBr + H slow iii H + Br  k3® HBr fast Solution:The fast equilibrium condition simply says thatk1 [Br2] = k-1[Br]2and [Br] = (k1/k-1 [Br2])½The slow step determines the rate law, rate = k2 [H2] [Br] Br is an intermediate = k2 [H2] (k1/k-1 [Br2])½ = k [H2] [Br2] ½; k = k2 (k1/k-1)½ M-½ s -1total order 1.5 explain 15 Chemical Kinetics

  39. Deriving a rate law from a mechanism - 3 The decomposition of N2O5 follows the mechanism: 1 N2O5 NO2 + NO3fast equilibrium 2 NO2 + NO3 —k2 NO + O2 + NO2slow 3 NO3 + NO —k3 NO2 + NO2fastDerive the rate law. Solution: The slow step determines the rate, rate = k2 [NO2] [NO3] NO2 & NO3 are intermediateFrom 1, we have [NO2] [NO3] —————— = KK, equilibrium constant [N2O5] K differ from k Thus, rate = K k2[N2O5] 15 Chemical Kinetics

  40. Deriving rate laws from mechanisms– steady-state approximation The steady-state approximation is a general method for deriving rate laws when the relative speed cannot be identified.It is based on the assumption that the concentration of the intermediate is constant. [Intermediate] Rate of producing the intermediate, Rprod, is the same as its rate of consumption, Rcons. Rprod = Rcons Rprod > Rcons Rprod < Rcons Be able to apply the steady-state approximation to derive rate laws time 15 Chemical Kinetics

  41. Let’s assume the mechanism for the reaction. H2 + I2 2 HIas follows.Step (1) I2—k1 2 IStep (1) 2 I —k-1 I2Step (2) H2 + 2 I —k2 2 HIDerive the rate law.Derivation:rate = k2 [H2] [I] 2 (‘cause this step gives products)but I is an intermediate, this is not a rate law yet.Since k1 [I2] (= rate of producing I) = k-1 [I]2 + k2 [H2] [I]2 (= rate of consuming I) Thus, k1 [I2] [I]2 = ——————k-1 + k2 [H2]  rate = k1k2 [H2] [I2] / {k-1 + k2 [H2] } Steady-state approximation - 2 Steady state 15 Chemical Kinetics

  42. Steady-state approximation - 3 From the previous result: k1k2 [H2] [I2]rate = ———————{k-1 + k2 [H2] } Discussion: (i) If k-1 << k2 [H2] then {k-1 + k2 [H2]} = k2 [H2] ,then rate = k1k2[H2] [I2] / {k2[H2] } = k1 [I2] (pseudo 1st order wrt I2)using large concentration of H2 or step 2 is fast (will meet this condition). (ii) If step (2) is slow, then k2 << k1, and if [H2] is not large, we have {k-1 + k2 [H2]} = k-1and rate = k1k2[H2] [I2] / k1 = k2[H2] [I2] 15 Chemical Kinetics

  43. Steady-state approximation - 4 In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O82- + SO32- + 2 OH- 3 SO42- + H2O.The following mechanism has been proposed: i S2O82- + SO32-—k1 S2O72- + SO42- ii S2O72- + H2O —k2 2 SO42- + 2 H+ iii H+ + OH-—k3 H2O (fast equilibrium to be discussed) Steady-state approximation follows these steps: What is or are the intermediates I?Use which step to give the rate law that may involve [I]? Express the rates of producing and consuming intermediate(s)Express [I] of intermediate(s) in terms of [Reactants] Derive the rate law in terms of [Reactants] Discuss See page 607 PHH Text 15 Chemical Kinetics

  44. Catalysis Energy A catalyst is a substance that changes the rate of a reaction by lowing the activation energy, Ea. It participates a reaction in forming an intermediate, but is regenerated. Enzymes are marvelously selective catalysts. A catalyzed reaction,NO (catalyst) 2 SO2 (g) + O2— 2 SO3 (g)via the mechanism i 2 NO + O2  2 NO2 (3rd order) ii NO2 + SO2  SO3 + NO Uncatalyzed rxn Catalyzed rxn rxn 15 Chemical Kinetics

  45. Catalyzed decomposition of ozone R.J. Plunkett in DuPont discovered carbon fluorine chlorine compounds. The CFC decomposes in the atmosphere:CFCl3® CFCl2 + ClCF2Cl3® CF2Cl + Cl. The Cl catalyzes the reaction via the mechanism: i O3 + h v® O + O2, ii ClO + O ® Cl + O2 iii O + O3® O2 + O2.The net result or reaction is2 O3® 3 O2 Scientists sound the alarm, and the CFC is banned now. 15 Chemical Kinetics

  46. Homogenous vs. heterogeneous catalysts A catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult. When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways. Platinum is often used to catalyze hydrogenation Catalytic converters reduce CO and NO emission. 15 Chemical Kinetics

  47. Heterogeneous catalysts Ceryx's vision is to design, produce, and commercialize advanced systems that balance Cost, Performance, Emissions Reduction, and Fuel Penalty to make the economics of pollution control viable.We explore new ways to look at the air quality challenges faced by industry and search for potential solutions by combining proven technologies with state-of-the-art science. Catalyzed reactions: CO + O2® CO22 NO ® N2 + O2 15 Chemical Kinetics

  48. 15 Chemical Kinetics

  49. Enzymes – selective catalysts Enzymes are a long protein molecules that fold into balls. They often have a metal coordinated to the O and N sites. Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site) of the enzyme molecules and just before and during the reaction. After having reacted, the products P1 & P2 are released. Enzyme + Substrate® ES (activated complex) ES ® P1 + P2 + E Enzymes are biological catalysts for biological systems. 15 Chemical Kinetics

  50. X-ray 3-D structure of fumarate reductase. It reduces fumerate, an important role in the metabolism of anaerobic bacteria, from Max Planck Inst. 15 Chemical Kinetics

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